POJ - 1753 Flip Game（状压枚举）

https://vjudge.net/problem/POJ-1753

题意

4*4的棋盘，翻转其中的一个棋子，会带动邻接的棋子一起动。现要求把所有棋子都翻成同一种颜色，问最少需要几步。

 

分析

同一个棋子翻偶数次等于没有翻，翻奇数次就浪费步数，因此每个棋子最多翻一次，也就是说，答案最大就是16。故总状态数就是2^16，可以直接dfs暴力。还有另一种思路就是状态压缩，把棋盘压成16位的数字，翻转时采用异或操作，我们暴力枚举每个状态，即所有选择棋子的可能情况跑一遍，对于每一个棋子，对其能影响的位置可以预处理出来，这样就通过位运算来模拟翻转过程了，更具体的看代码。

 

 

#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>

#define rep(i,e) for(int i=0;i<(e);i++)
#define rep1(i,e) for(int i=1;i<=(e);i++)
#define repx(i,x,e) for(int i=(x);i<=(e);i++)
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define mset(var,val) memset(var,val,sizeof(var))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pd(a) printf("%d\n",a)
#define scl(a) scanf("%lld",&a)
#define scll(a,b) scanf("%lld%lld",&a,&b)
#define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define IOS ios::sync_with_stdio(false);cin.tie(0)

using namespace std;
typedef long long ll;
template <class T>
void test(T a){cout<<a<<endl;}
template <class T,class T2>
void test(T a,T2 b){cout<<a<<" "<<b<<endl;}
template <class T,class T2,class T3>
void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;}
template <class T>
inline bool scan_d(T &ret){
    char c;int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret = ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
const int N = 1e6+10;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const ll mod = 1000000000;
int T;

void testcase(){
    printf("Case %d:",++T);
}

const int MAXN = 5e5+10 ;
const int MAXM = 150;
const double eps = 1e-8;
const double PI = acos(-1.0);
int n;
int st[16] = {0x13,0x27,0x4e,0x8c,0x131,0x272,0x4e4,0x8c8,0x1310,0x2720,0x4e40,0x8c80,0x3100,0x7200,0xe400,0xc800};
int po[16]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768};
int g[5][5];

void work(){
    char s[6];
    int state=0;
    for(int i=0;i<4;i++){
        scanf("%s",s);
        for(int j=0;j<4;j++){
            if(s[j]=='b') state += po[4*i+j];
        }
    }
    int ans=inf;
    for(int i=0;i<(1<<16);i++){
        int cnt = 0;
        int temp = state;
        for(int j=0;j<16;j++){
            if(i & po[j]){
                temp ^= st[j];
                cnt++;
            }
        }
        if(temp==0 || temp == 65535){
            if(ans>cnt){
                ans=cnt;
            }
        }

    }
    if(ans==inf) puts("Impossible");
    else cout<<ans<<endl;
    return;
}
int main() {
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
//    init();
    work();
    return 0;
}

 

深搜的做法，规定一定的搜索顺序，递归回溯。

#include<iostream>
#include<cmath>
#include<cstring>
#include<queue>
#include<vector>
#include<cstdio>
#include<algorithm>
#include<map>
#include<set>

#define rep(i,e) for(int i=0;i<(e);i++)
#define rep1(i,e) for(int i=1;i<=(e);i++)
#define repx(i,x,e) for(int i=(x);i<=(e);i++)
#define X first
#define Y second
#define PB push_back
#define MP make_pair
#define mset(var,val) memset(var,val,sizeof(var))
#define scd(a) scanf("%d",&a)
#define scdd(a,b) scanf("%d%d",&a,&b)
#define scddd(a,b,c) scanf("%d%d%d",&a,&b,&c)
#define pd(a) printf("%d\n",a)
#define scl(a) scanf("%lld",&a)
#define scll(a,b) scanf("%lld%lld",&a,&b)
#define sclll(a,b,c) scanf("%lld%lld%lld",&a,&b,&c)
#define IOS ios::sync_with_stdio(false);cin.tie(0)

using namespace std;
typedef long long ll;
template <class T>
void test(T a){cout<<a<<endl;}
template <class T,class T2>
void test(T a,T2 b){cout<<a<<" "<<b<<endl;}
template <class T,class T2,class T3>
void test(T a,T2 b,T3 c){cout<<a<<" "<<b<<" "<<c<<endl;}
template <class T>
inline bool scan_d(T &ret){
    char c;int sgn;
    if(c=getchar(),c==EOF) return 0;
    while(c!='-'&&(c<'0'||c>'9')) c=getchar();
    sgn=(c=='-')?-1:1;
    ret=(c=='-')?0:(c-'0');
    while(c=getchar(),c>='0'&&c<='9') ret = ret*10+(c-'0');
    ret*=sgn;
    return 1;
}
const int N = 1e6+10;
const int inf = 0x3f3f3f3f;
const ll INF = 0x3f3f3f3f3f3f3f3fll;
const ll mod = 1000000000;
int T;

void testcase(){
    printf("Case %d:",++T);
}

const int MAXN = 5e5+10 ;
const int MAXM = 150;
const double eps = 1e-8;
const double PI = acos(-1.0);
int n;
int g[5][5];
bool f;
bool check(){
    int t = g[0][0];
    for(int i=0;i<4;i++)
        for(int j=0;j<4;j++)
            if(t!=g[i][j])
                return false;
    return true;
}

void flip(int x,int y){
    g[x][y] = 1-g[x][y];
    if(x-1>=0) g[x-1][y] = 1-g[x-1][y];
    if(y-1>=0) g[x][y-1] = 1-g[x][y-1];
    if(x+1<4) g[x+1][y] = 1-g[x+1][y];
    if(y+1<4) g[x][y+1] = 1-g[x][y+1];
}

void dfs(int x,int y,int state){
    if(state==0){
        f = check();
        return;
    }
    if(f||y>3) return;
    flip(x,y);
    if(x<3) dfs(x+1,y,state-1);
    else dfs(0,y+1,state-1);
    flip(x,y);
    if(x<3) dfs(x+1,y,state);
    else dfs(0,y+1,state);
    return;
}
void work(){
    char s[6];
    f=false;
    for(int i=0;i<4;i++){
        scanf("%s",s);
        for(int j=0;j<4;j++){
            if(s[j]=='b') g[i][j]=0;
            else g[i][j]=1;
        }
    }

    for(n=0;n<=16;n++){
        dfs(0,0,n);
        if(f) break;
    }
    if(f) cout<<n<<endl;
    else puts("Impossible");
    return;
}
int main() {
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif // LOCAL
//    init();
    work();
    return 0;
}